limit

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canvas

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can you tell me guys if am I right?[math]\lim_{x\to0}x\sin\left(\frac{\pi}{x}\right)=\lim_{x\to0}x\frac{\sin\left(\frac{\pi}{x}\right)}{\frac{\pi}{x}}\frac{x}{\pi}=\pi\lim_{x\to0}\frac{\sin\left(\frac{\pi}{x}\right)}{\frac{\pi}{x}}\lim_{x\to0}x^2=\pi\cdot1\cdot0=0[/math]
 
canvas said:
can you tell me guys if am I right?[math]\lim_{x\to0}x\sin\left(\frac{\pi}{x}\right)=\lim_{x\to0}x\frac{\sin\left(\frac{\pi}{x}\right)}{\frac{\pi}{x}}\frac{x}{\pi}=\pi\lim_{x\to0}\frac{\sin\left(\frac{\pi}{x}\right)}{\frac{\pi}{x}}\lim_{x\to0}x^2=\pi\cdot1\cdot0=0[/math]
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Study this LINK and tell us what you think.
 
ok, I think that this is the mistake I've made:

[math]\lim_{x\to0}\frac{\sin(\square)}{\square}=1\,\,\Leftrightarrow\,\,\lim_{x\to0}\square=0[/math]
am I right?
 
canvas said:
ok, I think that this is the mistake I've made:
[math]\lim_{x\to0}\frac{\sin(\square)}{\square}=1\,\,\Leftrightarrow\,\,\lim_{x\to0}\square=0[/math]am I right?
Click to expand...
Actually NO. You have missed that [imath]\mathop {\lim }\limits_{x \to 0} \left( {\frac{\pi }{x}} \right)\ne 0[/imath]
[imath]\left| {x\sin \left( {\dfrac{\pi }{x}} \right)} \right| = \left| x \right|\left| {\sin \left( {\dfrac{\pi }{x}} \right)} \right| \leqslant \left| x \right|[/imath] Can you explain that?
What is the limit?
 
canvas said:
ok, I think that this is the mistake I've made:

[math]\lim_{x\to0}\frac{\sin(\square)}{\square}=1\,\,\Leftrightarrow\,\,\lim_{x\to0}\square=0[/math]
am I right?
Click to expand...
I am assuming that you are saying that the same expressions go into the first two boxes. OK, that is fine.
In \(\displaystyle \lim_{x\to0}\dfrac{sinx}{x}=1\), note that not only are the two x's exactly the same but they are both going to 0 as x goes to 0.
In \(\displaystyle \lim_{x\to0}\dfrac{sin2x}{2x}=1\), note that not only are the two 2x's exactly the same but they are both going to 0 as x goes to 0.
In \(\displaystyle \lim_{x\to5}\dfrac{sin(x-5)}{x-5}=1\), note that not only are the two (x-5)'s exactly the same but they are both going to 0 as x goes to 5
In \(\displaystyle \lim_{x\to\infty}\dfrac{sin(1/x)}{1/x}=1\), note that not only are the two (1/x)'s exactly the same but they are both going to 0 as x goes to \(\displaystyle \infty\).
 
canvas said:
can you tell me guys if am I right?[math]\lim_{x\to0}x\sin\left(\frac{\pi}{x}\right)=\lim_{x\to0}x\frac{\sin\left(\frac{\pi}{x}\right)}{\frac{\pi}{x}}\frac{x}{\pi}=\pi\lim_{x\to0}\frac{\sin\left(\frac{\pi}{x}\right)}{\frac{\pi}{x}}\lim_{x\to0}x^2=\pi\cdot1\cdot0=0[/math]
Click to expand...
One error is in your algebra. It is not true that [imath]\sin\left(\frac{\pi}{x}\right)=\frac{\sin\left(\frac{\pi}{x}\right)}{\frac{\pi}{x}}\frac{x}{\pi}[/imath]. Can you see the error?

Of course, the bigger error is what @pka pointed out, namely that [imath]\lim_{x\to0}\frac{\sin\left(\frac{\pi}{x}\right)}{\frac{\pi}{x}}\ne1[/imath] because [imath]\lim_{x\to0}\frac{\pi}{x}\ne0[/imath]
 
The reason why \(\displaystyle \lim_{x\to 0}\dfrac{sinx}{x} = 1\) is because sinx and x are extremely close to one another when x is near 0. As you know, when you divide two numbers that are close to one another you get a result close to 1.
As mentioned in my earlier post, not only do the two arguments have to be the same for the limit to be 1 but also the arguments have to approach 0 as x goes to whatever it is going to.
 
so it should be: [math]\lim_{x\to\triangle}\frac{\sin(\square)}{\square}=1\,\,\Leftrightarrow\,\,\lim_{x\to\triangle}\square=0[/math] right?
 
canvas said:
so it should be: [math]\lim_{x\to\triangle}\frac{\sin(\square)}{\square}=1\,\,\Leftrightarrow\,\,\lim_{x\to\triangle}\square=0[/math] right?
Click to expand...
You seem to be understanding what I said. Good!
 
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